Numbers $n$ such that the digit sums of $n, n^2,\cdots,n^k$ coincide.

The following probabilistic argument suggests that $k=3$ is very likely the maximum.

For $0 \leq \sigma \leq 9$ define $$ b(\sigma) = \min_{0 \leq x \leq +\infty} x^{-\sigma} \sum_{i=0}^9 x^i. $$ Thus $b(0)=b(9)=1$ and $b(9/2) = 10$.

Lemma. Suppose $0 \leq s \leq 9d$. Of the $10^d$ integers in $[0,10^d)$, at most $b(s/d)^d$ have digit sum $s$. The inequality is strict unless $s=0$ or $s=9d$.

Proof: Fix $d$, and let $N_s$ ($0 \leq s \leq 9d$) be the number of integers in $[0,10^d)$ with digit sum $s$. Then the $N_s$ have the generating polynomial $$ \sum_{s=0}^{9d} N_s X^s = \left( \sum_{i=0}^9 X^i \right)^d. $$ It follows that $$ N_s \leq x^{-s} \sum_{s'=0}^{9d} N_{s'} x^{s'} = x^{-s} \left( \sum_{i=0}^9 x^i \right)^d $$ for all $x>0$. Choosing the $x$ that minimizes this upper bound yields $N_s \leq b(s/d)^d$ as claimed, and it is clear that equality holds only for $s=0$ and $s=9d$. $\Box$

(Using contour integration one can show that this upper bound is within a factor $O(\sqrt d\,)$ of the actual count.)

Now for large $d$ it seems reasonable to expect that for random $n \in [0,10^d)$ the digit sums $S(n),S(n^2),\ldots,S(n^k)$ behave like random variables, independent except for the congruence mod $9$, with each $n^j$ behaving like a random $jd$-digit number. This means that the number of $d$-digit numbers $n$ for which $S(n)=S(n^2)=\cdots=S(n^k)$ is at most $B_k^d$, where $$ B_k = \max_{0 < \sigma < 9} b(\sigma) \prod_{j=2}^k \left( \frac{b(\sigma/j)}{10} \right)^j. $$ Numerical calculation finds $B_k \doteq 10, 6.56, 2.36, 0.39$ for $k=1,2,3,4$. Thus we expect infinitely many solutions for $k \leq 3$ but only finitely many for $k \geq 4$, and probably none now that Nikhil Mahajan has searched up to $k=8$ and found nothing. Indeed even $S(n) = S(n^4)$ should have only finitely many solutions (possibly none other than $n=7,19,67$) because even $\max_\sigma b(\sigma) \, (b(\sigma/4)/10)^4$ is about $0.855 < 1$. If you want to look for further examples, concentrate on $n \equiv 0,1,4,7 \bmod 9$ with $S(n)$ approximately equal to $7$ times the number of digits of $n$.

Some large examples of $S(n)=S(n^2)=S(n^3)$ are $n = 4659468895$, $n = 22898799351$, and $n=319879688698$.